**Dalton's atomic theory** (1808) proposed atoms as indivisible, ultimate particles of matter. However, experiments on electrical discharge through gases revealed that atoms contain smaller charged particles, fundamentally contradicting Dalton's model. The electrical nature of matter became apparent when substances like glass and ebonite developed charge when rubbed with silk or fur.
**Cathode Ray Discharge Tube Experiments:**
In the mid-1850s, scientists including **Michael Faraday** studied electrical discharge through partially evacuated glass tubes containing two metal electrodes (cathode and anode). When high voltage (thousands of volts) was applied at very low pressure, a stream of particles flowed from cathode to anode.
**Key Observations:**
**Conclusion:** These particles were named **electrons**, and their properties proved they are fundamental constituents of all atoms.
**Exam Point:** Cathode rays bend toward positive plate in electric field, confirming negative charge.
**J.J. Thomson's Experiment (1897):**
Thomson applied perpendicular electric and magnetic fields to cathode rays to measure the **e/m ratio** (charge-to-mass ratio).
**Method:**
**Factors affecting deflection:**
**Result:**
$$\frac{e}{m_e} = 1.758820 \times 10^{11} \text{ C kg}^{-1}$$
Where **e** = magnitude of electron charge, **m_e** = mass of electron
**Exam Note:** Higher charge-to-mass ratio means easier deflection in fields.
**R.A. Millikan's Oil Drop Experiment (1906-1914):**
**Principle:** Suspended charged oil droplets in electric field; determined charge by measuring electric field strength needed to balance gravitational force.
**Key Findings:**
**Result:**
**Mass of Electron:**
From **e/m_e ratio** and known charge:
$$m_e = 9.1094 \times 10^{-31} \text{ kg}$$
**Exam Calculation Example:**
If e/m ratio = 1.76 × 10¹¹ C/kg and e = 1.6 × 10⁻¹⁹ C, find mass:
$$m_e = \frac{1.6 \times 10^{-19}}{1.76 \times 10^{11}} = 9.09 \times 10^{-31} \text{ kg}$$
**Canal Rays (Positive Particles):**
Modified cathode ray tubes produced **canal rays** moving opposite to cathode rays, containing positively charged particles.
**Characteristics of Canal Rays:**
**Discovery of Proton:**
**Discovery of Neutron:**
**Table 2.1 Summary:**
| Particle | Charge (C) | Relative Charge | Mass (kg) | Mass (u) |
|----------|-----------|-----------------|-----------|----------|
| Electron | −1.602 × 10⁻¹⁹ | −1 | 9.109 × 10⁻³¹ | 0.00054 |
| Proton | +1.602 × 10⁻¹⁹ | +1 | 1.673 × 10⁻²⁷ | 1.00727 |
| Neutron | 0 | 0 | 1.675 × 10⁻²⁷ | 1.00867 |
**Exam Point:** Proton ≈ 1836 times heavier than electron; neutron ≈ 1839 times heavier than electron.
---
After discovering sub-atomic particles, scientists needed to explain:
**Proposed (1898):**
J.J. Thomson suggested atom is a **uniform sphere of positive charge** (radius ≈ 10⁻¹⁰ m) with **electrons embedded** throughout like plums in pudding or seeds in watermelon.
**Key Features:**
**Limitations:**
**Nobel Prize:** Thomson awarded 1906 for work on electricity conduction in gases.
**Rutherford's α-Particle Scattering Experiment:**
**Apparatus:**
**Expected Results (Thomson Model):**
**Actual Results:**
**Rutherford's Conclusions:**
1. **Most atom is empty space:** Since most particles passed undeflected, positive charge not uniformly distributed
2. **Positive charge concentrated:** Few particles deflected backward proved enormous repulsive force, so positive charge concentrated in tiny region called **nucleus**
3. **Nucleus dimensions:**
**Rutherford's Nuclear Model:**
**Mathematical Analysis:**
For α-particle deflection by nuclear charge:
**Limitations of Rutherford Model:**
**Atomic Number (Z):**
$$Z = \text{Number of protons} = \text{Number of electrons (in neutral atom)}$$
**Example:**
**Mass Number (A):**
$$A = Z + n = \text{Protons} + \text{Neutrons}$$
Where **n** = number of neutrons
**Example Calculations:**
Carbon-12: Z = 6, A = 12
Oxygen-16: Z = 8, A = 16
**Notation:** Element represented as $$^A_Z\text{X}$$
**Example:** $$^{12}_6\text{C}$$ = Carbon-12 (6 protons, 6 neutrons)
**Exam Point:** Atomic number determines element; mass number varies for same element.
**Isotopes:**
**Isobars:**
**Exam Distinction:**
**Calculation Example:**
Find composition of $$^{32}_{16}\text{S}$$:
Find mass number for atom with 26 protons, 30 neutrons:
---
**Particle Properties Established:**
**Model Evolution:**
**Atomic Composition:**
**Identification System:**
This foundation led to development of quantum mechanical model to explain stability and spectra—topics in subsequent sections.
Q1. Which of the following statements about cathode rays is NOT correct?
Answer: C — Cathode ray properties are independent of electrode material and gas type, proving electrons are universal constituents of all atoms.
Q2. The charge-to-mass ratio (e/me) of an electron is 1.758820 × 10¹¹ C kg⁻¹. If the charge on electron is –1.602176 × 10⁻¹⁹ C, what is the mass of electron?
Answer: A — me = e ÷ (e/me) = 1.602176 × 10⁻¹⁹ ÷ 1.758820 × 10¹¹ = 9.11 × 10⁻³¹ kg.
Q3. In Thomson's e/me measurement apparatus, the electrons hit point B on the screen when both electric and magnetic fields are applied. What does this indicate?
Answer: C — Point B is on the original path (no field), so electrons experience no net force; thus qE = qvB (forces are balanced).
Q4. Millikan's oil drop experiment directly measures which property of electron?
Answer: A — The oil drop experiment balances gravitational and electric forces on a suspended charged oil drop, allowing calculation of the electron's charge.
Q5. If the deflection of electrons in an electric field increases when the field strength increases, and decreases when electron mass increases, which statement best explains this observation?
Answer: B — From F = qE and a = F/m, deflection ∝ E/m, so deflection increases with field strength and decreases with mass.
Q6. A student observes that when a cathode ray tube is evacuated to different pressures, the properties of the rays remain unchanged. This observation proves that:
Answer: B — Since cathode ray properties are independent of gas type and pressure, electrons must be universal building blocks of all matter.
Q7. In Faraday's electrolysis experiments, chemical reactions occurred at the electrodes. This observation was significant because it suggested:
Answer: B — Discrete chemical reactions at specific electrodes suggested electricity consists of discrete particles, not continuous waves.
Q8. Consider two electrons: one deflected by 2 cm and another by 4 cm in the same electric field. Assuming both have the same charge, which statement is true?
Answer: B — Greater deflection (4 cm) indicates smaller mass; since deflection ∝ 1/m, the electron deflected 2 cm has twice the mass of the one deflected 4 cm.
Q9. Both Thomson's e/me measurement and Millikan's charge measurement are important. Which statement correctly relates these two experiments?
Answer: C — Thomson found e/me = 1.76 × 10¹¹ C/kg; Millikan found e = 1.6 × 10⁻¹⁹ C; combining them gives me = 9.11 × 10⁻³¹ kg.
Q10. A cathode ray tube experiment shows that when voltage across electrodes increases, electron deflection in a perpendicular electric field also increases. This is because: (A) Electrons gain higher velocity, so they spend less time in the field. (B) Higher voltage creates stronger deflecting field, so electrons experience greater force. (C) Electrons become heavier at higher voltages. (D) The relationship between voltage and deflection is inversely proportional.
Answer: B — Higher voltage between electrodes increases the electric field strength, which increases the force on electrons (F = qE), causing greater deflection.
What is the charge-to-mass ratio (e/me) of an electron determined by Thomson?
1.758820 × 10¹¹ C/kg, measured using crossed electric and magnetic fields in a cathode ray tube.
What is the magnitude of the charge on a single electron?
1.602176 × 10⁻¹⁹ C (or approximately 1.6 × 10⁻¹⁹ C), determined by Millikan's oil drop experiment.
What is the mass of an electron in kilograms?
9.1094 × 10⁻³¹ kg, calculated from e/me ratio and the known electron charge.
In a cathode ray tube, what do cathode rays consist of?
Negatively charged particles called electrons, which flow from the cathode (negative electrode) to the anode (positive electrode).
What is the key observation from Faraday's electrolysis experiments?
Chemical reactions at electrodes when electricity passes through electrolyte solutions, suggesting the particulate nature of electricity.
How did Thomson balance the deflection of electrons in crossed electric and magnetic fields?
By adjusting field strengths until electrons travelled straight without deflection, allowing calculation of the e/me ratio.
Why do cathode ray properties not depend on electrode material or gas type?
Because electrons are fundamental constituents of all atoms, so they are identical regardless of the substance used.
What does Millikan's oil drop experiment measure and how?
Measures the charge on an electron by balancing gravitational and electric forces on a suspended charged oil drop.
What happens to an electron's deflection in an electric field if its mass decreases?
The deflection increases because a lighter particle experiences greater acceleration from the same electric force.
Which scientist first showed that electricity has particulate nature through electrolysis?
Michael Faraday in 1830, who observed chemical reactions at electrodes and formulated laws of electrolysis.
Define the term 'electron' and state any two of its characteristic properties discovered through cathode ray experiments. [2 marks]
Electron is a negatively charged subatomic particle. State two from: (1) travels from cathode to anode, (2) properties independent of electrode/gas material, (3) deflected in electric/magnetic fields, (4) charge = –1.6 × 10⁻¹⁹ C.
Explain how Thomson determined the charge-to-mass ratio (e/me) of electron using a cathode ray tube with crossed electric and magnetic fields. Include a description of the key principle used to balance the forces. [5 marks]
Describe: (1) setup with perpendicular E and B fields, (2) deflection at point A (E-field only) and point C (B-field only), (3) balancing condition: qE = qvB at point B (straight path), (4) calculation of e/me from field strengths and electron velocity. Show working: e/me ∝ E/B².
Millikan's oil drop experiment is a landmark achievement in determining fundamental properties of matter. Explain the principle of the experiment, derive the condition for a stationary oil drop, and calculate the number of electrons on a drop if the charge on the drop is –3.2 × 10⁻¹⁹ C (given: charge on one electron = –1.6 × 10⁻¹⁹ C). [6 marks]
Explain: (1) setup with charged oil drop suspended between parallel metal plates, (2) forces acting: weight (mg) downward and electric force (qE) upward, (3) equilibrium condition: mg = qE, (4) from this, charge q can be found, (5) number of electrons = total charge ÷ charge per electron. Calculate: n = 3.2 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 2 electrons. Connect to why this proved quantization of charge.
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